One gram sample of `NH_(4)NO_(3)` is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by `6.12K` . The heat capacity of the system is `1.23KJ//g//deg` . What is the molar heat of decomposition for `NH_(4)NO_(3)` ?

Options

A. `-7.53 kJ//mol`
B. `-398.1 kJ//mol`
C. `-16.1 kJ//mol`
D. `-602 kJ//mol`

Explanation

To solve the problem, we need to determine the molar heat of decomposition for ammonium nitrate (NH₄NO₃) based on the provided data.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of NH₄NO₃ (m) = 1 gram
- Temperature change (ΔT) = 6.12 K
- Heat capacity of the system (C) = 1.23 kJ/g·°C

2. **Calculate the Heat (Q) Released:**
The heat released during the decomposition can be calculated using the formula:
\[ Q = m \times C \times \Delta T \]
Substituting the values:
\[ Q = 1 \, \text{g} \times 1.23 \, \text{kJ/g·°C} \times 6.12 \, \text{K} \]
\[ Q = 1 \times 1.23 \times 6.12 = 7.5276 \, \text{kJ} \]

3. **Calculate the Molar Mass of NH₄NO₃:**
The molar mass of NH₄NO₃ can be calculated as follows:
- Nitrogen (N): 14 g/mol × 2 = 28 g/mol
- Hydrogen (H): 1 g/mol × 4 = 4 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol
\[ \text{Molar mass of NH₄NO₃} = 28 + 4 + 48 = 80 \, \text{g/mol} \]

4. **Calculate the Molar Heat of Decomposition:**
Since we have calculated the heat released for 1 gram of NH₄NO₃, we can find the heat released for 80 grams (1 mole):
\[ \text{Molar heat of decomposition} = Q \times \frac{80 \, \text{g}}{1 \, \text{g}} = 7.5276 \, \text{kJ} \times 80 \]
\[ = 602.208 \, \text{kJ/mol} \]

5. **Determine the Sign of the Molar Heat:**
Since the decomposition is an exothermic reaction, the heat released will be negative:
\[ \text{Molar heat of decomposition} = -602.208 \, \text{kJ/mol} \approx -602 \, \text{kJ/mol} \]

### Final Answer:
The molar heat of decomposition for NH₄NO₃ is approximately **-602 kJ/mol**.